Suppose $$ x_1, x_2, x_3, \dots, x_{100} $$ are in arithmetic progression such that $$ x_5 = -4 $$ and $$ 2x_6 + 2x_9 = x_{11} + x_{13} $$. Then, $$ x_{100} $$ equals:

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Accepted Answer

The problem involves finding the value of $$x_{100}$$ in an arithmetic progression (AP) given certain conditions. Here's the step-by-step solution:

**General Form of AP:**

- The terms of an AP are expressed as:
     $$
     x_n = x_1 + (n-1)d
     $$
     where $$x_1$$ is the first term, $$d$$ is the common difference, and $$n$$ is the term number.

**Given Conditions:**

- $$x_5 = -4$$
   - $$2x_6 + 2x_9 = x_{11} + x_{13}$$

**Express Terms Using AP Formula:**

- From $$x_5 = -4$$, we know:
     $$
     x_5 = x_1 + 4d \quad 	ext{so} \quad x_1 = -4 - 4d
     $$
   - Substitute into other terms:
     $$
     x_6 = x_1 + 5d = (-4 - 4d) + 5d = -4 + d
     $$
     $$
     x_9 = x_1 + 8d = (-4 - 4d) + 8d = -4 + 4d
     $$
     $$
     x_{11} = x_1 + 10d = (-4 - 4d) + 10d = -4 + 6d
     $$
     $$
     x_{13} = x_1 + 12d = (-4 - 4d) + 12d = -4 + 8d
     $$

**Use the Equation:**

Substitute into $$2x_6 + 2x_9 = x_{11} + x_{13}$$:
   $$
   2(-4 + d) + 2(-4 + 4d) = (-4 + 6d) + (-4 + 8d)
   $$
   Simplify:
   $$
   -8 + 2d -8 + 8d = -8 + 6d -8 + 8d
   $$
   Combine terms:
   $$
   -16 + 10d = -16 + 14d
   $$
   Solve for $$d$$:
   $$
   10d = 14d \quad \Rightarrow \quad d = -2
   $$

**Find $$x_{100}$$:**

Using the formula for the nth term:
   $$
   x_{100} = x_1 + (100-1)d
   $$
   Substitute values:
   $$
   x_{100} = (-4 - 4(-2)) + (99)(-2)
   $$
   Simplify:
   $$
   x_{100} = (-4 + 8) - 198
   $$
   $$
   x_{100} = -194
   $$

### Final Answer:
The value of $$x_{100}$$ is **-194**.  
Correct option: **B**

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